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Infinite Series help needed

  • Thread starter sobek
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  • #1
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http://www4a.wolframalpha.com/Calculate/MSP/MSP167199e5bhg1gg5673i000048ed16i8cbf5iacg?MSPStoreType=image/gif&s=14&w=256&h=40 [Broken]

for all x in the interval of convergence of the given power series.

a) write the first 3 nonzero terms and the general term for an infinite series that represents http://www4a.wolframalpha.com/Calculate/MSP/MSP34199e5dhc325d3hd200005ag14201e835f640?MSPStoreType=image/gif&s=52&w=70&h=36 [Broken]

b) Find the sum of the series as determined in part a.

I have no idea how to do these two parts. I just don't know where to begin. Any help is appreciated thanks.
 
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Answers and Replies

  • #2
rock.freak667
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Your general term is

[tex]\frac{(n+1)x^n}{3^{n+1}}[/tex]

so your sum can be represented as

[tex]f(x) = \sum_{n=0} ^N \frac{(n+1)x^n}{3^{n+1}}[/tex]

Now when you integrate within the summation sign, you just integrate what is being summed

e.g.

[tex] \int \sum_{n=0} ^N f(x) dx = \sum_{n=0} ^N \int f(x) dx[/tex]
 
  • #3
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Okay so I can just take the integral of the general term and find the sum of the first 3 nonzero terms?
 
  • #4
rock.freak667
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Okay so I can just take the integral of the general term and find the sum of the first 3 nonzero terms?
Yes, then you can write out the first three terms and easily find a closed form for the sum.
 
  • #5
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Ok, I understand that, but what exactly are the limits of integration (0,1) there for?
 
  • #6
rock.freak667
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Ok, I understand that, but what exactly are the limits of integration (0,1) there for?
Because you will need to integrate between those limits

[tex]\sum_{n=0} ^N \int_0 ^1 f(x) dx[/tex]
 
  • #7
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Agh! Ok I'm confused. If I'm just taking the integral of the general form, do I still need those limits? Or am I doing a Fundamental Theorem thing?
 
  • #8
rock.freak667
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Agh! Ok I'm confused. If I'm just taking the integral of the general form, do I still need those limits? Or am I doing a Fundamental Theorem thing?
yes you would still need those limits. It would be equivalent of doing the integral between those limits of every term.


Thus you just need to compute

[tex]\int_0 ^1 \frac{(n+1)x^n}{3^{n+1}}dx [/tex]
 
  • #9
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Okay so how exactly would I integrate this? Seeing as there are two variables.
 
  • #10
rock.freak667
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Okay so how exactly would I integrate this? Seeing as there are two variables.
Treat 'n' as a constant.
 
  • #11
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Ok so for the integral I got http://www4a.wolframalpha.com/Calculate/MSP/MSP758199e8ha121cf0a4c00004ggc11fi1gf83a0h?MSPStoreType=image/gif&s=29&w=34&h=40 [Broken]

Is that right?

and from that is the sum: infinity?
 
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  • #12
rock.freak667
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Ok so for the integral I got http://www4a.wolframalpha.com/Calculate/MSP/MSP758199e8ha121cf0a4c00004ggc11fi1gf83a0h?MSPStoreType=image/gif&s=29&w=34&h=40 [Broken]

Is that right?

and from that is the sum: infinity?
substitute the limits of the integral.


But now your sum will go from n=-1 to ∞
 
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  • #13
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Okay with the limits of integration substituted I got http://www2.wolframalpha.com/Calculate/MSP/MSP344199e931ddgafc99d0000507h04h0277efcda?MSPStoreType=image/gif&s=9&w=34&h=37

and then i find the first 3 nozero terms from that right?
 
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  • #14
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Now, I can't see the original question (the link to wolframalpha is dead), but this doesn't sound right.

Can you post the problem again, sobek?
 
  • #15
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http://www2.wolframalpha.com/Calculate/MSP/MSP608199e96816fc40d8a000022aig2805fea6cad?MSPStoreType=image/gif&s=10&w=65&h=38

for all x in the interval of convergence of the given power series.

a) write the first 3 nonzero terms and the general term for an infinite series that represents http://www2.wolframalpha.com/Calculate/MSP/MSP179199e974h5cca8ib200004hf8af3ab6c4i576?MSPStoreType=image/gif&s=12&w=70&h=36

b) Find the sum of the series as determined in part a.

I'm still having trouble on part A
 
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  • #16
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Okay, I see it now.

Okay with the limits of integration substituted I got http://www2.wolframalpha.com/Calculate/MSP/MSP344199e931ddgafc99d0000507h04h0277efcda?MSPStoreType=image/gif&s=9&w=34&h=37

and then i find the first 3 nozero terms from that right?
Yes.

Assuming your original power series started at n=0, start from there.

Let n=0 in [tex]\frac 1{3^{n+1}}[/tex] to find the first term. Then let n=1 to find your next, etc.
 
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  • #17
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Okay so my math in getting https://www.physicsforums.com/latex_images/26/2602939-0.png [Broken] was right?

and because it's a geometric series I can use a/1-r to find the sum right?
 
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  • #18
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Yes and yes.
 
  • #19
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Thanks to rocknerd667 and mathnerdmo for helping me with this problem
 

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